IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl IV. Three-Phase Induction Machines Induction Machines 1 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 2 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 3 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 4 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 5 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 6 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 7 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 8 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 9 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 10 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 11 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 12 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 13 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Example 1: A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters: R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Calculate the following information: i. ii. iii. iv. v. vi. vii. viii. Synchronous speed in rpm slip Line Current Input Power Airgap Power Torque Developed Output Power in Hp Efficiency This machine has no iron loss resistance, so the equivalent circuit is as follows: i. Synchronous speed is given by: ii. Slip is given by Using the rpm equation, s = (1200-1170)/1200 = 0.025 iii. Now, phase current is given by 14 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl where phase impedance is given by Using the above equation, Z in = 9.57 + j3.84 Ω And noting that the machine is delta connected, V 1 = V LL = 480V I 1 = 43.1 - j17.4 A. |I 1 | =46.6 A, θ = -21.9° Therefore I L = √3 × 46.6 = 80.6 A iv. Input power is given by: P in = 62.2 kW v. Airgap power is the input power minus stator losses. In this case the core losses are grouped with rotational loss. Therefore P gap = 62.2 kW - 3× 46.62 × 0.461 P gap = 59.2 kW vi. Torque developed can be found from where synchronous speed in radians per second is given by giving τ = 471 Nm Vii.Output power in horsepower is the output power in Watts divided by 746. (there are 746 W in one Hp). and 15 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Therefore output power in Watts is: P out = 55.3kW Viii. Efficiency is given by η = 55.3/62.2 = 88.9% 16 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 17 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 18 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 19 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 20 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Using the induction machine power and torque equations it is possible to produce the torque speed curve shown below. Operating Regions The torque-speed curve brakes down into three operating regions: 1. Braking, n m < 0, s > 1 Torque is positive whilst speed is negative. Considering the power conversion equation it can be seen that if the power converted is negative (from P = τ ω) then the airgap power is positive. i.e. the power is flowing from the stator to the rotor and also into the rotor from the mechanical system. This operation is also called plugging. This mode of operation can be used to quickly stop a machine. If a motor is travelling forwards it can be stopped by interchanging the connections to two of the three phases. Switching two phases has the result of changing the direction of motion of the stator magnetic field, effectively putting the machine into braking mode in the opposite direction. 2. Motoring, 0 < n m < n s , 1 > s > 0 Torque and motion are in the same direction. This is the most common mode of operation. 21 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 3. Generating, n m > n s , s < 0 In this mode, again torque is positive whilst speed is negative. However, unlike plugging, indicates that if the power converted is negative, so is the air gap power. In this case, power flows from the mechanical system, to the rotor circuit, then across the air gap to the stator circuit and external electrical system. Motoring Torque Characteristic The motoring region of the induction machine torque-speed curve is the region of greatest interest. • Starting torque – Torque at zero speed – Typically 1.5 times the full-load torque • Pull-up torque – The minimum torque developed by the motor while accelerating from zero speed – Greater than full-load torque; less than starting torque • Breakdown torque – The maximum torque that the motor can develop – Typically 2.5 times the full-load torque • Normal operation – At full-load, the motor runs at n rpm – Rotor speed decreases slightly from synchronous speed with increasing load torque – Motor will stall when the load torque exceeds the breakdown torque 22 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl The torque equation Using the equation Multiple solutions of the above equation for torque at different slips can be made simpler by simplifying the equivalent circuit model. Consider the diagram below: The stator part of the equivalent circuit (together with the magnetising branch) can be replaced by a Thevenin equivalent circuit. In the Thevenin circuit, the stator phase voltage has been replaced by its Thevenin equivalent, 23 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl and the impedances have been replaced by Thevenin equivalent impedances. Incorporating the Thevenin model into the circuit model results in the Thevenin equivalent circuit model of an induction machine. In the above circuit, the calculation of rotor current is greatly simplified The above expression for rotor current can be squared and substituted into the torque equation Using the above equation, the variation of torque with slip can be plotted directly . Note that if power or efficiency calculations are needed, the full equivalent circuit model should be used (not the Thevenin version). since synchonous speed is constant, maximum torque occurs at the same slip as maximum airgap power. Considering the Thevenin circuit, and applying maximum power transfer theory, maximum airgap power and maximum torque will occur when π π 2 2 = οΏ½π π ππππ + (ππππππ + ππ2 )2 π π 24 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Re-arranging it is possible to obtain the slip for maxiumum torque, or pullout torque. ππππππππ = π π 2 2 οΏ½π π ππππ + (ππππππ + ππ2 )2 Substituting the pullout slip into the Thevenin torque equation: ππππππππ = 2 3ππππππ 2πππ π π π π π π π οΏ½π π ππππ + οΏ½π π 2ππππ + (ππππππ + ππ2 )2 οΏ½ The Starting torque • • • • 2 The starting torque is proportional to ππππππ πππ π π π π π π π π π = ππ 2 3ππππππ π π 2 2 +(ππ 2 [(π π ) +π π π π π π π π π π ππππ 2 ππππ +ππ2 ) ] The starting torque will be reduced if the stator and rotor leakage inductances are increased The starting torque will be reduced if the stator frequency is increased When the rotor resistance is increased, the starting torque will first increase and then decrease. πππ π π π π π π π π π ππππππ = ππππππππ when π π 2= ππ1 + ππ2 25 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Equivalent Circuit Model Analysis Example A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor has the following parameters: R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω Rotational losses are 2450W. The motor drives a mechanical load at a speed of 1170 rpm. Find: i. ii. iii. iv. Thevenin circuit parameters and Thevenin voltage Pullout slip Pullout Torque Start Torque Solution: Using Matlab or Excel (or another computer program) plot the torque speed curve for slip in the range 0 to 1 i. Thevenin circuit parameters and Thevenin voltage: The Thevenin voltage is the voltage applied to the rotor assuming that the rotor current is zero. Thevenin impedance is the impedance of the stator part of the circuit, seen from the rotor, assuming that the stator supply is short circuited. i. Substituting the equivalent circuit parameters in to the above equations gives: V TH = 475.2 V, R TH = 0.452Ω, X TH = 0.313Ω i. Pullout slip The slip at which maximum torque occurs can be found from maximum power transfer theory. Maximum torque and maximum airgap power occur at the same slip, therefore maximum torque occurs when 26 IV. Three-phase Induction Machines ii. Dr. Suad Ibrahim Shahl Pullout Torque Pullout torque can be found by substituting the above pullout slip into the Thevenin torque equation or from the maximum torque equation directly Substituting into the above equation: i. Start Torque Start torque can be found by setting s=1 in the above equation for torque. 27 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Effect of Rotor Resistance 28 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Investigating the torque-speed curve it is apparent that the rotor circuit resistance has significant impact on speed at which maximum torque occurs. The plots below illustrate two cases, with low rotor resistance on the left and high rotor resistance on the right. Example 1: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following impedances in ohms per phase referred to the stator circuit: πΉπΉππ = ππ. ππππππππ πΏπΏππ = ππ. ππππππππ πΉπΉππ = ππ. ππππππππ πΏπΏππ = ππ. ππππππππ πΏπΏπ΄π΄ = ππππ. ππππ a) What is the maximum torque of this motor? At what speed and slip does it occur? b) What is the starting torque of this motor? c) When the rotor resistance is doubled, what is the speed at which the maximum torque now occurs? What is the new starting torque of the motor? Solution: The Thevenin voltage of this motor is The Thevenin resistance is The Thevenin reactance is ππππππ = ππππ = 2 ππππ οΏ½ 1 +ππ ππ π π ππππ ≈ π π 1 οΏ½ ππ ππππβ ≈ ππ1 ≈ 1.106ππ οΏ½π π 12 +(ππ1 +ππππ )2 (266 ππ)(26.3 ππ ) οΏ½(ππ.ππππππππ )2 +(ππ.ππππππππ +ππππ.ππππ )2 2 ππππ.ππππ οΏ½ ππ.ππππππππ+ππππ.ππππ ≈ (ππ. ππππππππ) οΏ½ ππππ = 255.2ππ = 0.590ππ (a) The slip at which maximum torque occurs is given by Equation (2): ππππππππ = π π 2 2 +(ππ 2 οΏ½π π ππππ ππππ +ππ2 ) = ππ.ππππππππ οΏ½(0.590ππ )2 +(1.106ππ+ππ.ππππππππ )2 = 0.198 This corresponds to a mechanical speed of ππππ = (1 − π π )πππ π π π π π π π β = (1 − 0.198)(1800 ππ/ππππππ) = 1444 ππ⁄ππππππ 29 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl The torque at this speed is ππππππππ = 2 3ππππππ 2 +(ππ 2 2ππ π π π π π π π π οΏ½π π ππππ +οΏ½π π ππππ ππππ +ππ2 ) οΏ½ 3(255.2 ππ)2 = 2(188.5 ππππππ /π π )[(0.590ππ)2 +(1.106ππ+ππ.ππππππππ)2 ] = 229 ππ. ππ (b) The starting torque of this motor is found by setting s=1 in Equation (1) : πππ π π π π π π π π π = = 2 3ππππππ π π 2 2 πππ π π π π π π π [(π π πππ»π» + π π 2 ) + (ππππππ + ππ2 )2 ] 3(255.2 ππ)2 (ππ. ππππππππ) = 104 ππ. ππ (188.5 ππππππ⁄π π )[(0.590ππ + ππ. ππππππππ)2 + (1.106ππ + ππ. ππππππππ)2 ] (c ) If the rotor resistance is doubled, then the slip at maximum torque doubles, too. Therefore, π π ππππππ = 0.396 and the speed at maximum torque is ππππ = (1 − π π )πππ π π π π π π π = (1 − 0.396)(1800 ππ⁄ππππππ) = 1087 ππ⁄ππππππ The maximum torque is still ππππππππ = 229 ππ. ππ The starting torque is now πππ π π π π π π π π π = 3(255.2ππ)2 (0.664ππ) = 170ππ. ππ (188.5 ππππππ⁄π π )[(0.590ππ + ππ. ππππππππ)2 + (1.106ππ + ππ. ππππππππ)2 ] 30 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 31 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 32 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl When the switch S is in the START position, the stator windings are connected in STAR When the motor picks up speed, say 80% of its rated value, the switch S is thrown quickly to the RUN position which connects the stator windings in DELTA. In induction motors that are designed to operate with delta stator connection it is possible, during starting, to reduce the phase voltage by switching to Y- connection During Y- connection, the phase voltage V s becomes so the phase current, for same slip, IsY , is reduced 3 times 33 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Now the line current in Δ connection I lΔ is So the line current is three times smaller for Y- connection. The torque is proportional to phase voltage squared Therefore, the Y-delta starting is equivalent to in torque. a reduction of phase voltage and a 3 to 1 reduction 34 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 35 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 36 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 37 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 38 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 39 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 40 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 41 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 42 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 43 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl 44 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Fig.1 Slip ring IM Fig.2 Torque-slip curves 45 IV. Three-phase Induction Machines Dr. Suad Ibrahim Shahl Fig.3 Torque-speed curves 46
